Find the probability of reaching all points after N moves from point N
Last Updated :
29 Oct, 2023
Given N which denotes the initial position of the person on the number line. Also given L which is the probability of the person of going left. Find the probability of reaching all points on the number line after N moves from point N. Each move can be either to the left or to the right.
Examples:
Input: n = 2, l = 0.5
Output: 0.2500 0.0000 0.5000 0.0000 0.2500
The person can't reach n-1th position and n+1th position in 2 passes, hence the probability is 0. The person can reach 0th position by only moving 2 steps left from index 2, hence the probability of reaching 0th index is 05*0.5=0.25. Similarly for 2n index, the probability is 0.25.
Input: n = 3, l = 0.1
Output: 0.0010 0.0000 0.0270 0.0000 0.2430 0.0000 0.7290
The person can reach n-1th in three ways, i.e., (llr, lrl, rll) where l denotes left and r denotes right. Hence the probability of n-1th index is 0.027. Similarly probabilities for all other points are also calculated.
Approach: Construct an array arr[n+1][2n+1] where each row represents a pass and the columns represent the points on the line. The maximum a person can move from index N is to 0th index at left or to 2nth index at right. Initially the probabilities after one pass will be left for arr[1][n-1] and right for arr[1][n+1]. The n-1 moves which are left will be done, hence the two possible moves will either be n steps to the right or n steps to the left. So the recurrence relations for right and left moves for all will be:
arr[i][j] += (arr[i - 1][j - 1] * right)
arr[i][j] += (arr[i - 1][j + 1] * left)
The summation of probabilities for all possible moves for any index will be stored in arr[n][i].
Below is the implementation of the above approach:
C++
// C++ program to calculate the
// probability of reaching all points
// after N moves from point N
#include <bits/stdc++.h>
using namespace std;
// Function to calculate the probabilities
void printProbabilities(int n, double left)
{
double right = 1 - left;
// Array where row represent the pass and the
// column represents the points on the line
double arr[n + 1][2 * n + 1] = {{0}};
// Initially the person can reach left
// or right with one move
arr[1][n + 1] = right;
arr[1][n - 1] = left;
// Calculate probabilities for N-1 moves
for (int i = 2; i <= n; i++)
{
// when the person moves from ith index in
// right direction when i moves has been done
for (int j = 1; j <= 2 * n; j++)
arr[i][j] += (arr[i - 1][j - 1] * right);
// when the person moves from ith index in
// left direction when i moves has been done
for (int j = 2 * n - 1; j >= 0; j--)
arr[i][j] += (arr[i - 1][j + 1] * left);
}
// Print the arr
for (int i = 0; i < 2*n+1; i++)
printf("%5.4f ", arr[n][i]);
}
// Driver Code
int main()
{
int n = 2;
double left = 0.5;
printProbabilities(n, left);
return 0;
}
/* This code is contributed by SujanDutta */
// Java program to calculate the
// probability of reaching all points
// after N moves from point N
import java.util.*;
class GFG {
// Function to calculate the probabilities
static void printProbabilities(int n, double left)
{
double right = 1 - left;
// Array where row represent the pass and the
// column represents the points on the line
double[][] arr = new double[n + 1][2 * n + 1];
// Initially the person can reach left
// or right with one move
arr[1][n + 1] = right;
arr[1][n - 1] = left;
// Calculate probabilities for N-1 moves
for (int i = 2; i <= n; i++) {
// when the person moves from ith index in
// right direction when i moves has been done
for (int j = 1; j <= 2 * n; j++) {
arr[i][j] += (arr[i - 1][j - 1] * right);
}
// when the person moves from ith index in
// left direction when i moves has been done
for (int j = 2 * n - 1; j >= 0; j--) {
arr[i][j] += (arr[i - 1][j + 1] * left);
}
}
// Calling function to print the array with probabilities
printArray(arr, n);
}
// Function that prints the array
static void printArray(double[][] arr, int n)
{
for (int i = 0; i < arr[0].length; i++) {
System.out.printf("%5.4f ", arr[n][i]);
}
}
// Driver Code
public static void main(String[] args)
{
int n = 2;
double left = 0.5;
printProbabilities(n, left);
}
}
# Python3 program to calculate the
# probability of reaching all points
# after N moves from point N
# Function to calculate the probabilities
def printProbabilities(n, left):
right = 1 - left;
# Array where row represent the pass
# and the column represents the
# points on the line
arr = [[0 for j in range(2 * n + 1)]
for i in range(n + 1)]
# Initially the person can reach
# left or right with one move
arr[1][n + 1] = right;
arr[1][n - 1] = left;
# Calculate probabilities
# for N-1 moves
for i in range(2, n + 1):
# When the person moves from ith
# index in right direction when i
# moves has been done
for j in range(1, 2 * n + 1):
arr[i][j] += (arr[i - 1][j - 1] * right);
# When the person moves from ith
# index in left direction when i
# moves has been done
for j in range(2 * n - 1, -1, -1):
arr[i][j] += (arr[i - 1][j + 1] * left);
# Print the arr
for i in range(2 * n + 1):
print("{:5.4f} ".format(arr[n][i]), end = ' ');
# Driver code
if __name__=="__main__":
n = 2;
left = 0.5;
printProbabilities(n, left);
# This code is contributed by rutvik_56
// C# program to calculate the
// probability of reaching all points
// after N moves from point N
using System;
class GFG
{
// Function to calculate the probabilities
static void printProbabilities(int n, double left)
{
double right = 1 - left;
// Array where row represent the pass and the
// column represents the points on the line
double[,] arr = new double[n + 1,2 * n + 1];
// Initially the person can reach left
// or right with one move
arr[1,n + 1] = right;
arr[1,n - 1] = left;
// Calculate probabilities for N-1 moves
for (int i = 2; i <= n; i++)
{
// when the person moves from ith index in
// right direction when i moves has been done
for (int j = 1; j <= 2 * n; j++)
{
arr[i, j] += (arr[i - 1, j - 1] * right);
}
// when the person moves from ith index in
// left direction when i moves has been done
for (int j = 2 * n - 1; j >= 0; j--)
{
arr[i, j] += (arr[i - 1, j + 1] * left);
}
}
// Calling function to print the array with probabilities
printArray(arr, n);
}
// Function that prints the array
static void printArray(double[,] arr, int n)
{
for (int i = 0; i < GetRow(arr,0).GetLength(0); i++)
{
Console.Write("{0:F4} ", arr[n,i]);
}
}
public static double[] GetRow(double[,] matrix, int row)
{
var rowLength = matrix.GetLength(1);
var rowVector = new double[rowLength];
for (var i = 0; i < rowLength; i++)
rowVector[i] = matrix[row, i];
return rowVector;
}
// Driver Code
public static void Main(String[] args)
{
int n = 2;
double left = 0.5;
printProbabilities(n, left);
}
}
/* This code contributed by PrinciRaj1992 */
// Function to calculate the probabilities
function printProbabilities(n, left) {
let right = 1 - left;
// Array where row represent the pass and the
// column represents the points on the line
let arr = new Array(n + 1);
for (let i = 0; i < n + 1; i++) {
arr[i] = new Array(2 * n + 1).fill(0);
}
// Initially the person can reach left
// or right with one move
arr[1][n + 1] = right;
arr[1][n - 1] = left;
// Calculate probabilities for N-1 moves
for (let i = 2; i <= n; i++)
{
// when the person moves from ith index in
// right direction when i moves has been done
for (let j = 1; j <= 2 * n; j++) {
arr[i][j] += (arr[i - 1][j - 1] * right);
}
// when the person moves from ith index in
// left direction when i moves has been done
for (let j = 2 * n - 1; j >= 0; j--) {
arr[i][j] += (arr[i - 1][j + 1] * left);
}
}
// Print the arr
for (let i = 0; i < 2*n+1; i++) {
console.log(`${arr[n][i].toFixed(4)} `);
}
}
// Driver Code
let n = 2;
let left = 0.5;
printProbabilities(n, left);
// This code is contributed by lokeshpotta20.
Output0.2500 0.0000 0.5000 0.0000 0.2500
Time Complexity: O(N2)
Auxiliary Space: O(N2)
Efficient approach : Space optimization
In previous approach the arr[i][j] is depend upon the current and previous row of 2D matrix. So to optimize space we use two vectors curr and dp that keep track of current and previous row of arr.
Implementation Steps:
- Initialize a vectors arr of size 2*N+1 to keep track of only previous row of arr matrix with 0.
- Initialize base case for the condition when person can reach left or right in one move.
- Now iterative over subproblems and get the current computation.
- While Initialize a vectors temp of size 2*N+1 to keep track of only current row of arr matrix with 0.
- At last traverse and print all values of arr.
Implementations Steps:
C++
#include <bits/stdc++.h>
using namespace std;
void printProbabilities(int n, double left)
{
double right = 1 - left;
// Array to store the probabilities
vector<double> arr(2 * n + 1, 0);
// Initially the person can reach left
// or right with one move
arr[n + 1] = right;
arr[n - 1] = left;
// Calculate probabilities for N-1 moves
for (int i = 2; i <= n; i++)
{
// Temp array to store updated probabilities
vector<double> temp(2 * n + 1, 0);
// when the person moves from ith index in
// right direction when i moves have been done
for (int j = 1; j <= 2 * n; j++)
temp[j] += (arr[j - 1] * right);
// when the person moves from ith index in
// left direction when i moves have been done
for (int j = 2 * n - 1; j >= 0; j--)
temp[j] += (arr[j + 1] * left);
// Copy the updated probabilities to the main array
arr = temp;
}
// Print the probabilities
for (int i = 0; i < 2 * n + 1; i++)
printf("%5.4f", arr[i]);
}
int main()
{
int n = 2;
double left = 0.5;
printProbabilities(n, left);
return 0;
}
import java.io.*;
import java.util.Arrays;
public class GFG {
// Function to calculate and
// print probabilities
static void printProbabilities(int n, double left) {
double right = 1 - left;
// Array to store the probabilities
double[] arr = new double[2 * n + 1];
// Initially the person can reach left or right with one move
arr[n + 1] = right;
arr[n - 1] = left;
// Calculate probabilities for N-1 moves
for (int i = 2; i <= n; i++) {
// Temp array to store updated probabilities
double[] temp = new double[2 * n + 1];
// when the person moves from ith index in
// right direction when i moves have been done
for (int j = 1; j <= 2 * n; j++)
temp[j] += (arr[j - 1] * right);
for (int j = 2 * n - 1; j >= 0; j--)
temp[j] += (arr[j + 1] * left);
// Copy the updated probabilities to
// the main array
arr = Arrays.copyOf(temp, temp.length);
}
// Print the probabilities
for (int i = 0; i < 2 * n + 1; i++)
System.out.printf("%5.4f ", arr[i]);
}
public static void main(String[] args) {
int n = 2;
double left = 0.5;
printProbabilities(n, left);
}
}
def print_probabilities(n, left):
right = 1 - left
# Array to store the probabilities
arr = [0] * (2 * n + 1)
# Initially the person can reach left
# or right with one move
arr[n + 1] = right
arr[n - 1] = left
# Calculate probabilities for N-1 moves
for i in range(2, n + 1):
# Temp array to store updated probabilities
temp = [0] * (2 * n + 1)
# when the person moves from ith index in
# right direction when i moves have been done
for j in range(1, 2 * n + 1):
temp[j] += (arr[j - 1] * right)
# when the person moves from ith index in
# left direction when i moves have been done
for j in range(2 * n - 1, -1, -1):
temp[j] += (arr[j + 1] * left)
# Copy the updated probabilities to the main array
arr = temp
# Print the probabilities
for i in range(2 * n + 1):
print(f"{arr[i]:.4f}", end=" ")
if __name__ == "__main__":
n = 2
left = 0.5
print_probabilities(n, left)
using System;
class GFG
{
static void PrintProbabilities(int n, double left)
{
double right = 1 - left;
// Array to store the probabilities
double[] arr = new double[2 * n + 1];
// Initially the person can reach left
// or right with one move
arr[n + 1] = right;
arr[n - 1] = left;
// Calculate probabilities for N-1 moves
for (int i = 2; i <= n; i++)
{
// Temp array to store updated probabilities
double[] temp = new double[2 * n + 1];
// when the person moves from ith index in
// right direction when i moves have been done
for (int j = 1; j <= 2 * n; j++)
temp[j] += (arr[j - 1] * right);
// when the person moves from ith index in
// left direction when i moves have been done
for (int j = 2 * n - 1; j >= 0; j--)
temp[j] += (arr[j + 1] * left);
// Copy the updated probabilities to the main array
arr = temp;
}
// Print the probabilities
for (int i = 0; i < 2 * n + 1; i++)
Console.Write(String.Format("{0:F4} ", arr[i]));
}
static void Main(string[] args)
{
int n = 2;
double left = 0.5;
PrintProbabilities(n, left);
}
}
function printProbabilities(n, left) {
let right = 1 - left;
// Array to store the probabilities
let arr = new Array(2 * n + 1).fill(0);
// Initially the person can reach left
// or right with one move
arr[n + 1] = right;
arr[n - 1] = left;
// Calculate probabilities for N-1 moves
for (let i = 2; i <= n; i++) {
// Temp array to store updated probabilities
let temp = new Array(2 * n + 1).fill(0);
// when the person moves from ith index in
// right direction when i moves have been done
for (let j = 1; j <= 2 * n; j++)
temp[j] += arr[j - 1] * right;
// when the person moves from ith index in
// left direction when i moves have been done
for (let j = 2 * n - 1; j >= 0; j--)
temp[j] += arr[j + 1] * left;
// Copy the updated probabilities to the main array
arr = [...temp];
}
// Print the probabilities
for (let i = 0; i < 2 * n + 1; i++)
console.log(arr[i].toFixed(4));
}
let n = 2;
let left = 0.5;
printProbabilities(n, left);
Output:
0.2500 0.0000 0.5000 0.0000 0.2500
Time Complexity: O(N^2)
Auxiliary Space: O(N)
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